Exercise 11.1

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°

Solution:

We use the formula for the area of the sector of a circle.

The formula for the area of the sector of a circle with radius 'r' and angle θ = (θ360°) × πr2

Given, θ = °, Radius = cm

Area of the sector = (θ360°) × πr2

= 60°360° × × 6 ×

= cm2

The area of a sector is 1327 cm2.

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

Let the radius of the circle be 'r'

Thus, the circumference C = 2πrπr2

Therefore, r = C2π

A quadrant of a circle means one of the four equal parts.

Therefore, the area of a quadrant = × Area of a circle = 14 × πr2

Circumference (C) = 22 cm

Therefore, radius (r) = C2π

r = 22(2 × 22/7)

r = (22 × 7)(2 × 22)

r = cm

Therefore, the area of a quadrant = 14 × πr2

= 14 × 227 × 72 ×

= cm2

Therefore, the area of a quadrant = 778 cm2

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

We use the formula for the area of the circle to find the area swept by the minute hand.

We know that the minute hand completes one rotation in 1 hour or minutes.

Therefore, area swept by the minute hand in 60 minutes = Area of the circle with radius equal to the length of the minute hand = πr2πr

Using unitary method, we get

Area swept by minute hand in 1 minute = πr260πr230

Thus, area swept by minute hand in 5 minutes = (πr260) × = πr212πr210

Length of the minute hand (r) = cm

It is known that the minute hand completes one rotation in 1 hour or 60 minutes

Therefore, the area swept by the minute hand in 60 minutes = πr2πr3

Therefore, the area swept by the minute hand in 5 minutes = × πr2

= πr2125πr212

= × 227 × × cm2

= cm2

The area swept by the minute hand in 5 minutes is equal to 1543 cm2.

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)

Solution:

We use the formula for the area of a sector of the circle to solve the problem.

In a circle with radius r and the angle at the centre with degree measure θ,

(i) Area of the sector = θ360° × πr2πr3

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle

Area of the right triangle = × base × height

Let's draw a figure to visualize the area to be calculated.

Here, Radius, r = cm, θ = °

Visually it’s clear from the figure that,

AB is the chord that subtends a right angle at the centre.

(i) Area of minor segment APB = Area of sector OAPB - Area of right triangle AOB

(ii) Area of major segment AQB = πr2 - Area of minor segment APB

Area of the right triangle ΔAOB = × OA ×

(i) Area of minor segment APB = Area of sector OAPB - Area of right ΔAOB

= θ360° × πr2 - 12 × × OB

= 90°360° × πr2 - 12 × ×

= πr24πr23 - r22r24

= r2×π4−12

= r2×3.14−24

= r2×1.144

= 10×10×1.144 cm2 (Since radius r is given as 10 cm)

= cm2

Thus, area of minor segment APB is equal to 28.5 cm2.

(ii) Area of major sector AOBQ = πr2 - Area of minor sector OAPB

= πr2 - θ360° × πr2

= πr21−90°360°

= × 10cm2 ×

= cm2

Thus, area of major segment AQB is equal to 235.5 cm2.

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.

Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

Solution:

In the circle with radius r and the angle at the centre with a degree measure of θ,

(i) Length of the Arc = θ360° × 2πrπr2

(ii) Area of the sector = θ360° × πr22πr

(iii) Area of the segment = Area of the sector - Area of the corresponding triangle

Let's draw a figure to visualize the problem.

Here, r = cm, θ = °

Visually it’s clear from the figure that,

Area of the segment APB = Area of sector AOPB - Area of ΔAOB

(i) Length of the Arc, APB = θ360° × 2πr3πr

= 60°360° × 2 × × 21 cm

= cm

(ii) Area of the sector, AOBP = θ360° x πr2πr3

= 60°360° × 227 × × cm2

= cm2

(iii) Area of the segment = Area of the sector AOBPABPO - Area of the triangle AOBBOP

To find the area of the segment, we need to find the area of ΔAOB

In ΔAOB, draw OM ⊥ .

Consider ΔOAM and ΔOMB,

OA = (radiichord of the circle)

OM = ( side)

∠OMA = ∠OMB = ° (Since OM AB)

Therefore, ΔOMB ≅ ΔOMA (By Congruency)

So, AM = (Corresponding parts of triangles are always equal)

∠AOM = ∠BOM = × ° = °

In ΔAOM,

cos 30° = OMOAOAOM and sin 30° = AMOAOAAM

3212 = OMr and = AMr

OM = (32) r and AM = (12) r

AB = AM

AB = 2 × (12) r

AB =

Therefore, area of ΔAOB = × AB ×

= 12 × × (32) r

= 12 × cm × (32) × cm

= 441×34440×34 cm2

Area of the segment formed by the chord = Area of the sector AOBP - Area of the triangle AOB

= - 4413444034 cm2

Thus, area of the segment formed by the chord is equal to (231 - 44134) cm2.

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.(Use π = 3.14 and 3 = 1.73)

Solution:

In a circle with radius r and the angle at the centre of degree measure θ,

(i) Area of the sector = θ360° × πr2πr3

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle

Let's draw a figure to visualize the area to be calculated.

Here, radius, r = cm, θ = °

Visually it’s clear from the figure that,

AB is the chord that subtends ° angle at the centre.

(i) Area of minor segment APB = Area of sector OAPB - Area of ΔAOBΔAPB

(ii) Area of major segment AQB = πr2πr3 - Area of minor segment APB

Here, radius, r = cm, θ = °

Area of the sector OAPB = θ360° × πr2

= 60°360° × × × cm2

= cm2 (Round off to two decimal places)

In ΔAOB,

OA = = r (radii of the circle)

∠OBA = ∠ (Angles opposite to the equal sides in a triangle are equal)

∠AOB + ∠OBA + ∠OAB = ° (Angle sum property of a triangle)

° + ∠OAB + ∠OAB = 180°

∠OAB = °

∠OAB = °

∴ ΔAOB is an triangle because all its angles are equal.

AB = OAAB = OB =

Area of ΔAOB = 34 × side2side

= 34 r2

= 34 × 15cm2

= 1.734 × cm2

= cm2 (Round off to two decimal places)

(i) Area of minor segment APB = Area of sector OAPB - Area of ΔAOB

= cm2 - cm2 (Round off to two decimal places)

= cm2 (Round off to two decimal places)

(ii) Area of the major segment AQB = Area of the circle - Area of minor segment APB

= π × 15cm2 - 20.44 cm2

= 3.14 × cm2 - 20.44 cm2

= cm2 - 20.44 cm2

= cm2 (Round off to two decimal places)

Area of the major segment AQB is equal to 686.06 cm2.

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle (Use π = 3.14 and 3 = 1.73)

Solution:

We use the concept of areas of sectors of circles to solve the question.

In a circle with radius r and the angle at the centre with degree measure θ:

(i) Area of the sector = θ360° πr2πr

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle

Let's draw a figure to visualize the given question.

Here, radius, r = cm, ∠AOB = θ = °

Visually it’s clear from the figure that AB is the chord that subtends 120° angle at the centre.

To find the area of the segment AYB, we have to find the area of the sector OAYB and the area of the ΔAOB

(i) Area of sector OAYB = θ360° πr2πr

(ii) Area of ΔAOB = 12 × base × height

For finding the area of ΔAOB, draw OM ⊥ AB then find base AB and height OM using the figure as shown above.

Area of sector OAYB = 120°360° × π r2

= × × 12cm2

= cm2

Draw a perpendicular OM from O to chord AB

In ΔAOM and ΔBOM

AO = BO = (radii of circle)

OM = (common side)

∠OMA = ∠OMB = ° (perpendicular OM drawn)

∴ ΔAOM ≅ ΔBOM (By Congruency Rule)

⇒ ∠AOM = ∠ (By CPCT)

Therefore, ∠AOM = ∠BOM = ∠AOB = °

In ΔAOM,

AMOA = sin ° and OMOAOAOM = cos 60°

AM12 = 3212 and OM12 =

AM = 32 × cm and OM = 12 × cm

AM = 6333 cm and OM = cm

⇒ AB = AM

= 2 × 63 cm

= 12383 cm

Area of ΔAOB = 12 × AB ×

= 12 × 123 cm × cm

= × 1.73 cm2

= cm2 (Round off to two decimal places)

Area of segment AYB = Area of sector OAYB - Area of ΔAOB

= cm2 - cm2 (Round off to two decimal places)

= cm2 (Round off to two decimal places)

Area of segment AYB is equal to 88.44 cm2.

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope see the below figure. Find

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Solution:

We use the concept of area of a sector and area of a square to solve the problem.

(i) From the figure it’s clear that the horse can graze an area of a sector of a circle with radius (r) 5m and an angle with a degree measure ° (Since its a field)

Length of the rope = m

Area of the field the horse can graze = Area of the sector with θ = ° and r =

= θ360° × πr2πr3

= 90°360° × πr2

= × π × 5m2

= 254 × m2

= m2 (Round off to two decimal places)

The area of that part of the field in which the horse can graze is 19.63 m2.

(ii) The area that can be grazed by the horse when the length of the rope is 10m, is the area of the sector of a circle with a radius (r) 10 m and an angle with a degree measure 90°.

Area of the field the horse can graze = Area of the sector with θ = 90° and r = 10

= 90°360° × πr22πr

= × × m2

= m2 (Round off to one decimal place)

Increase in the grazing area

= m2 - m2

= m2

The increase in the grazing area is 58.87 m2.

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in below figure.

Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Solution:

(i) Diameter of the brooch (d) = mm

Total length of silver wire required = circumference of brooch + × diameter

= πdπr + 5d

= (π + ) ×

= ( + 5) × 35

= 22+357 × 35

= 57 × 5

= mm

Total length of the silver wire required is 285 mm.

(ii) Radius of the brooch (r) = mm

The wire divides the brooch into equal sectors.

So, angle of the sector (θ) = 360°1010360° = °

∴ Area of each sector of the brooch = 36°360° × πr2πr

= × 227 × mm × mm

= mm2

Area of each sector of the brooch is equal to 96.25 mm2.

10. An umbrella has 8 ribs which are equally spaced see the below figure. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

We will use the concepts related to the area of a sector of a circle to solve the problem.

There are 8 equally spaced ribs in an umbrella, so the umbrella is assumed to be a flat .

∴ The angle between 2 consecutive ribs (θ) = 360°88360° = °

The radius of the flat circle (r) is given as cm.

The area between 2 consecutive ribs of the umbrella = Area of a sector with an angle of 45° = θ360° × πr2πr

= 45°360° × × cm × cm

= × 227 × 45 cm × 45 cm

= 22275283237516 cm2

= cm2 (Round off to two decimal places)

The area between the two consecutive ribs of the umbrella is 795.54 cm2.

11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution:

Let's draw a figure according to the given question.

Visually it is clear that the area cleaned at the sweep of blades of each wiper = Area of the sector with angle ° at the center and radius of the circle cm.

There are 2 wipers of the same blade length and same angle of sweeping. Also, there is no area of overlap for the wipers.

∴ Total area cleaned at each sweep of the blades = × Area cleaned at the sweep of each wiper.

Area cleaned at the sweep of each wiper = Area of the sector of a circle with radius cm at an angle of °

= θ360° × πr2πr

= 115°360° × π × ×

= × π

There are two identical blade-length wipers.

∴ Total area cleaned at each sweep of the blades = 2 × 2372 × 625π

= 2 × 2372 × 227 × 625

= 23×11×62518×7

= 158125126158325126 cm2

= cm2 (Round off to two decimal places)

Total area cleaned at each sweep of the blades is equal to 1254.96 cm2.

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.(Use π = 3.14)

Solution:

We use the area of a sector of a circle to solve the given question.

The lighthouse spreads red coloured light over a sector of a circle with a radius km and an angle of degree measure °.

Area of the sea over which the ships are warned = Area of the sector of the circle with radius 16.5 km with an angle of degree measure 80°

= θ360° × πr2πr

= 80°360° × πr2

= πr2

= 29 × × km × km

= km2 (Round off to two decimal places)

Area of the sea over which the ships are warned is 189.97 km2.

13. A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm2. (Use 3 = 1.7)

Solution:

We know that, in a circle with radius r and the angle at the center with degree measure θ,

(i) Area of the sector = θ360 × πr2πr

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle

Visually it is clear from the diagram that the designs are segments of the circle.

∴ Area of the design = Area of segments of the circle.

Since the table cover has 6 equal design, therefore angle of each sector at the center = 360°6 = °

Consider segment APB. Chord AB subtends an angle of 60° at the center.

∴ Area of segment APB = Area of sector AOPB - Area of ΔAOB

Consider ΔAOB:

OB = (radii of the circle)

∠OAB = ∠ (angles to equal sides in a triangle are )

∠AOB + ∠OAB + ∠OBA = ° (angle sum property of a triangle)

∠OAB = ° - ° (Since, ∠AOB = 60°)

∠OAB = 120°2 = °

∴ ΔAOB is an triangle.

Area of ΔAOB = 34side2

= 34282 (Since the side of the triangle = radii of the circle = 28 cm)

= 3 × × 28

= 19631953

= 196 × 1.7

= cm2 (Round off upto one decimal place)

Area of sector OAPB = 60°360° × πr2

= × × ×

= 11×4×283

= cm2

Area of segment APB = Area of sector OAPB - Area of ΔAOB

= (12323 - 333.2) cm2

Area of the designs = × Area of segment

= 6 × (12323 - 333.2) cm2

= - cm2

= cm2 (Round off to one decimal place)

Cost of making 1 cm2 of designs = ₹

∴ Cost of making 464.8 cm2 of designs

= ₹ 0.35 ×

= ₹ (Round off to two decimal places)

Cost of making the designs is ₹ 162.68.

14. Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is ?

We use the concept of the area of sectors of a circle to solve the problem.

Consider, area of the sector of angle θ = θ360° × πr2πr, where r is the radius of the circle

Here, θ = and r =

Substituting the above values in the formula, we get the area of the sector = p360°2p360° × πR2πr2

Multiplying numerator and denominator of p360° × πR2 by 2, we get

Hence, Area of the sector = p720°2p720° × 2πR2